Increasing and Decreasing Functions and Asymptote Analysis

Vocabulary flashcards covering domain, asymptotes, derivatives, and intervals of increase and decrease for rational functions from Lesson 5.

Domain of $f(x) = \frac{x^2 - 3x - 4}{x - 2}$

The set of all real numbers excluding $x = 2$, expressed in interval notation as $(-\infty, 2) \cup (2, \infty)$.

Vertical Asymptote of $f(x) = \frac{x^2 - 3x - 4}{x - 2}$

The vertical line $x = 2$, determined where the one-sided limits approach infinite values: $\lim_{x \to 2^-} f(x) = +\infty$ and $\lim_{x \to 2^+} f(x) = -\infty$.

Condition for a Slant Asymptote (S.A.)

Occurs in a rational function when the degree of the numerator $n$ is exactly one greater than the degree of the denominator $m$, satisfying $n = m + 1$.

Slant Asymptote of $f(x) = \frac{x^2 - 3x - 4}{x - 2}$

The linear function $y = x - 1$, obtained via polynomial long division yielding $f(x) = x - 1 + \frac{-6}{x - 2}$ where $\lim_{x \to \pm\infty} [f(x) - (x - 1)] = 0$.

First Derivative of $f(x) = \frac{x^2 - 3x - 4}{x - 2}$

$f'(x) = \frac{x^2 - 4x + 10}{(x - 2)^2}$, which has no real roots in the numerator and remains strictly positive for all points in its domain.

Intervals of Increase for $f(x) = \frac{x^2 - 3x - 4}{x - 2}$

The function is increasing on $(-\infty, 2) \cup (2, \infty)$ because $f'(x) > 0$ across its entire domain.

Domain of $f(x) = \frac{x^2}{x^2 - 9}$

The set of real numbers excluding $x = -3$ and $x = 3$, represented in interval notation as $(-\infty, -3) \cup (-3, 3) \cup (3, +\infty)$.

Vertical Asymptotes of $f(x) = \frac{x^2}{x^2 - 9}$

The lines $x = -3$ and $x = 3$, where the denominator equals zero and function values tend toward $\pm\infty$.

Horizontal Asymptote of $f(x) = \frac{x^2}{x^2 - 9}$

The line $y = 1$, determined by evaluating the limit as $x$ approaches infinity: $\lim_{x \to \pm\infty} \frac{x^2}{x^2 - 9} = 1$.

First Derivative of $f(x) = \frac{x^2}{x^2 - 9}$

$f'(x) = \frac{-18x}{(x^2 - 9)^2}$, derived using the quotient rule.

Critical Number of $f(x) = \frac{x^2}{x^2 - 9}$

$x = 0$, found by setting the numerator of the derivative $-18x = 0$.

Intervals of Increase for $f(x) = \frac{x^2}{x^2 - 9}$

The intervals where $f'(x) > 0$, given as $(-\infty, -3) \cup (-3, 0)$.

Intervals of Decrease for $f(x) = \frac{x^2}{x^2 - 9}$

The intervals where $f'(x) < 0$, given as $(0, 3) \cup (3, +\infty)$.